Question

您好我是java函数式编程的新手。我试图过滤一个地图，如果元素在集合中，跳过它;否则处理它并将其放入集合中。但我发现它看起来像保持静态（不是由终端操作更新）

public class StreamMap {

    public static void main(String[] args) {

        Map<Integer, String> HOSTING = new HashMap<>();
        HOSTING.put(1, "linode.com");
        HOSTING.put(2, "heroku.com");
        HOSTING.put(3, "digitalocean.com");
        HOSTING.put(4, "aws.amazon.com");
        HOSTING.put(5, "aws.amazon.com");

        Set<String> s = new HashSet();

        //Map -> Stream -> Filter -> Map
        HOSTING.entrySet().stream()
                .filter(map -> !s.contains(map.getValue()))
                .collect(Collectors.toMap(p -> p.getKey(), p -> p.getValue())).forEach((k, v) -> {System.out.println("show:" + String.valueOf(k)  + v);s.add(v);});


    }
}

返回

show:linode.com
show:heroku.com
show:digitalocean.com
show:aws.amazon.com
show:aws.amazon.com //I don't want this to be returns because it's already done before

Answer 1

尝试一下谎言：

    HOSTING.entrySet()
            .stream()
            .filter(map -> s.add(map.getValue()))
            .collect(Collectors.toMap(p -> p.getKey(), p -> p.getValue())).forEach((k, v) -> {
        System.out.println("show:" + v);
    });

Answer 2

另一种方法是您可以将地图值转换为列表，然后您可以轻松地在列表中使用 Distinct（）

 List<String> result = HOSTING.values().stream().collect(Collectors.toList()); //convert map to list
 result.stream().distinct().collect(Collectors.toList());  // filtering the list

 System.out.println(result);

java过滤动态集

2 个答案: